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A certain soil has the following properties: Gs = 2.71, n = 40% and w = 20%. The degree of saturation of the soil (rounded off to the nearest percent) is 

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Concept:

Degree of saturation (S): 

It is the ratio of volume of water to the volume of voids present in soil sample. Its value vary from 0 to 1.

\({\rm{S = }}\frac{{{{\rm{V}}_{\rm{w}}}}}{{{{\rm{V}}_{\rm{V}}}}}\)

Void ratio (e): 

It is the ratio of volume of voids to the volume of solids present in soil sample. Void ratio of fine grained soid is greater than coarse grained soil.

\({\rm{e = }}\frac{{{{\rm{V}}_{\rm{V}}}}}{{{{\rm{V}}_{\rm{S}}}}}\)

Water content (w): 

It is the ratio of weight of water to the weight of solids present in soil sample. Water content of fine grained soid is greater than coarse grained soil.

\({\rm{w = }}\frac{{{{\rm{W}}_{\rm{w}}}}}{{{{\rm{W}}_{\rm{s}}}}}\)

Specific gravity (G): 

It is the ratio of unit weight of any substance to the unit weight of water.

\({\rm{G = }}\frac{{{{\rm{\gamma }}_{\rm{s}}}}}{{{{\rm{\gamma }}_{\rm{w}}}}}\)

The relationship between e and n is given by,

\({\rm{\;n}} = {\rm{}}\frac{{\rm{e}}}{{1\: + \:{\rm{e}}}}\)

The relationship between S, e, w and G is given by,

Se = wG

Calculation:

Given,

Gs = 2.71, n = 40% = 0.4 and w = 20% = 0.2

The relationship between e and n is given by

\({\rm{n}} = {\rm{}}\frac{{\rm{e}}}{{1\: + \:{\rm{e}}}}\)

It can be written as,

\(\begin{array}{l} {\rm{e}} = \frac{{\rm{n }}}{{1 - {\rm{n }}}} \end{array}\)

\(\begin{array}{l} {\rm{e}} = \frac{{\rm{0.4 }}}{{1 - {\rm{0.4 }}}} \end{array}\)

e = 0.67

The relationship between S, e, w and G is given by

Se = wG

S × 0.67 = 0.2 × 2.71

S = 0.808 = 80.8% ≈ 81%

Hence the degree of saturation of the soil  is 81%

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